• #### what is solution of y''* y'-y(x)^n=0 ?

From Dr Huang@21:1/5 to All on Fri Dec 10 20:12:18 2021
what is solution of y''* y'-y(x)^n=0 ?
wolfram solution is wrong ? There are bugs in many software such as WolframAlpha. Wish wolfram fix the bugs. You can copy equation and paste into MathHand.com, click the ODE button to solve, then click the test button to test its solution.

over 500 bugs in
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• From Nasser M. Abbasi@21:1/5 to Dr Huang on Sat Dec 11 01:48:39 2021
On 12/10/2021 10:12 PM, Dr Huang wrote:
what is solution of y''* y'-y(x)^n=0 ?
wolfram solution is wrong ? There are bugs in many software such as WolframAlpha. Wish wolfram fix the bugs. You can copy equation and paste into MathHand.com, click the ODE button to solve, then click the test button to test its solution.

over 500 bugs in
DrHuang.com/index/bugs

This can be solved by transformation on the dependent variable using v(y)=y'(x).

It results in a first order ode in v(y). The solution is then found,
then a new first order ode in y(x) now result which is solved again.
Both are separable.

But the integrals are tricky to do . I used Maple for the integrations.

Here is my solution which Maple's verifies. (there are 3 solutions)

https://12000.org/tmp/121121/foo2.pdf

https://12000.org/tmp/121121/foo2.htm

Maple 2021.2 gives these 3 solutions (Maple can't integrate the separable odes)

Intat(1/(-1/2/(1+n)*((3*_a^(1+n)-_C1)*(1+n)^2)^(1/3)-1/2*I*3^(1/2)/(1+n)*((3*_a ^(1+n)-_C1)*(1+n)^2)^(1/3)),_a = y(x))-x-_C2 = 0, Intat(1/(-1/2/(1+n)*((3*_a^(1 +n)-_C1)*(1+n)^2)^(1/3)+1/2*I*3^(1/2)/(1+n)*((3*_a^(1+n)-_C1)*(1+n)^2)^(1/3)), _a = y(x))-x-_C2 = 0, Intat((1+n)/((3*_a^(1+n)-_C1)*(1+n)^2)^(1/3),_a = y(x))-x -_C2 = 0

Mathematica 13 can integrate the separable odes' and gives these solutions
in terms of Hypergeometric functions

{{y[x]->InverseFunction[-(((-1/3)^(1/3)*(1+n)^(1/3)*Hypergeometric2F1 [1/3,(1+n)^(-1),1+(1+n)^(-1),-(#1^(1+n)/((1+n)*C[1]))]*#1 *(1+#1^(1+n)/((1+n)*C[1]))^(1/3))/((1+n)*C[1]+#1^(1+n))^(1/3))&] [x+C[2]]},{y[x]->InverseFunction[((1+n)^(1/3)*Hypergeometric2F1 [1/3,(1+n)^(-1),1+(1+n)^(-1),-(#1^(1+n)/((1+n)*C[1]))]*#1*(1+ #1^(1+n)/((1+n)*C[1]))^(1/3))/(3^(1/3)*((1+n)*C[1]+#1^(1+n))^(1/3))&] [x+C[2]]},{y[x]->InverseFunction[((-1)^(2/3)*(1+n)^(1/3) *Hypergeometric2F1[1/3,(1+n)^(-1),1+(1+n)^(-1),-(#1^(1+n)/((1+n)*C[1]))] *#1*(1+#1^(1+n)/((1+n)*C[1]))^(1/3))/(3^(1/3)*((1+n)*C[1]+#1^(1+n))^(1/3))&][x+C[2]]}}

--Nasser

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• From Dr Huang@21:1/5 to Nasser M. Abbasi on Thu Mar 10 16:06:27 2022
On Saturday, 11 December 2021 at 18:48:41 UTC+11, Nasser M. Abbasi wrote:
On 12/10/2021 10:12 PM, Dr Huang wrote:
what is solution of y''* y'-y(x)^n=0 ?
wolfram solution is wrong ? There are bugs in many software such as WolframAlpha. Wish wolfram fix the bugs. You can copy equation and paste into MathHand.com, click the ODE button to solve, then click the test button to test its solution.

over 500 bugs in
DrHuang.com/index/bugs

This can be solved by transformation on the dependent variable using v(y)=y'(x).

It results in a first order ode in v(y). The solution is then found,
then a new first order ode in y(x) now result which is solved again.
Both are separable.

But the integrals are tricky to do . I used Maple for the integrations.

Here is my solution which Maple's verifies. (there are 3 solutions)

https://12000.org/tmp/121121/foo2.pdf

https://12000.org/tmp/121121/foo2.htm

Maple 2021.2 gives these 3 solutions (Maple can't integrate the separable odes)

Intat(1/(-1/2/(1+n)*((3*_a^(1+n)-_C1)*(1+n)^2)^(1/3)-1/2*I*3^(1/2)/(1+n)*((3*_a
^(1+n)-_C1)*(1+n)^2)^(1/3)),_a = y(x))-x-_C2 = 0, Intat(1/(-1/2/(1+n)*((3*_a^(1
+n)-_C1)*(1+n)^2)^(1/3)+1/2*I*3^(1/2)/(1+n)*((3*_a^(1+n)-_C1)*(1+n)^2)^(1/3)),
_a = y(x))-x-_C2 = 0, Intat((1+n)/((3*_a^(1+n)-_C1)*(1+n)^2)^(1/3),_a = y(x))-x
-_C2 = 0

Mathematica 13 can integrate the separable odes' and gives these solutions
in terms of Hypergeometric functions

{{y[x]->InverseFunction[-(((-1/3)^(1/3)*(1+n)^(1/3)*Hypergeometric2F1 [1/3,(1+n)^(-1),1+(1+n)^(-1),-(#1^(1+n)/((1+n)*C[1]))]*#1 *(1+#1^(1+n)/((1+n)*C[1]))^(1/3))/((1+n)*C[1]+#1^(1+n))^(1/3))&] [x+C[2]]},{y[x]->InverseFunction[((1+n)^(1/3)*Hypergeometric2F1 [1/3,(1+n)^(-1),1+(1+n)^(-1),-(#1^(1+n)/((1+n)*C[1]))]*#1*(1+ #1^(1+n)/((1+n)*C[1]))^(1/3))/(3^(1/3)*((1+n)*C[1]+#1^(1+n))^(1/3))&] [x+C[2]]},{y[x]->InverseFunction[((-1)^(2/3)*(1+n)^(1/3) *Hypergeometric2F1[1/3,(1+n)^(-1),1+(1+n)^(-1),-(#1^(1+n)/((1+n)*C[1]))] *#1*(1+#1^(1+n)/((1+n)*C[1]))^(1/3))/(3^(1/3)*((1+n)*C[1]+#1^(1+n))^(1/3))&][x+C[2]]}}

--Nasser

thanks,t hey are too complicated. input your ODE into mathHand.com, click the ODE button to solve, then the test button to test its solution. or click the link:
http://server.drhuang.com/input/?guess=ODE%28y%27%27*+y%27-y%5En%29&inp=y%27%27*+y%27-y%5En&lang=null

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